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il63 [147K]
3 years ago
15

(Atwood’s Machine): Two masses, 9 kg and 12 kg, are attached by a lightweight cord and suspended over a frictionless pulley. Whe

n released, find the acceleration of the system and the tension in the cord.

Physics
2 answers:
Fittoniya [83]3 years ago
5 0

Answer:

Acceleration = 1.428m/s2

Tension = 102.85N

Explanation:

The detailed solution is attached

USPshnik [31]3 years ago
5 0

Answer:

The acceleration of the system is 1.401 m/s² and

The tension in the cord is 100.902 N

Explanation:

Let the 9 kg mass be m

Let the 12 kg mass be M

By Newton's second law of motion we have

For the 9 kg mass, T - mg = ma and for the 12 kg mass we have T - Mg  = -Ma

Here we took the upward acceleration as positive a of the 9 kg mass and the downward acceleration of the 12 kg mass as -a

Solving for T for the 9 kg mass we have

T = mg + ma

Substituting  the value of T in to the 12 kg mass equation, we have

mg + ma - Mg = -Ma or  a = (\frac{M-m}{M+m} )g therefore the acceleration is

1.401 m/s²

and the tension is T = mg + ma = 9×(9.81+1.401) = 100.902 N

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James and John dive from an overhang into the lake below. James simply drops straight down from the edge. John takes a running s
liraira [26]

Answer:

Both of them reach the lake at the same time.

Explanation:

We have equation of motion s = ut + 0.5at²

Vertical motion of James : -

          Initial velocity, u = 0 m/s

         Acceleration, a = g

         Displacement, s = h

    Substituting,

                  s = ut + 0.5 at²

                 h = 0 x t + 0.5 x g x t²

                 t_{James}=\sqrt{\frac{2h}{g}}

Vertical motion of John : -

          Initial velocity, u = 0 m/s

         Acceleration, a = g

         Displacement, s = h

    Substituting,

                  s = ut + 0.5 at²

                 h = 0 x t + 0.5 x g x t²

                 t_{John}=\sqrt{\frac{2h}{g}}

So both times are same.

Both of them reach the lake at the same time.

3 0
3 years ago
Use the drop-down menus to complete the statement.
Nikitich [7]
Classics.

Resistance is equal to relation between voltage and current.

R = \frac{U}{I}

If we express current:

I = \frac{U}{R}

If current is in fact 0 then one of the quantities either voltage or resistance must be equal to zero. Since resistance cannot be equal 0, because that would violate mathematical law that states that division by zero is undefined the only logical conclusion is voltage.

So the answer should be C voltage and B zero.

Hope this helps!
8 0
3 years ago
Read 2 more answers
A mass weighting 48 lbs stretches a spring 6 inches. The mass is in a medium that exerts a viscous resistance of 27 lbs when the
Mademuasel [1]

Answer:

a)

u(t)=0.499ft.e^{-\frac{144.76lb/s}{2(48lb)}t}cos(\omega t)\\\\u(t)=0.499ft.e^{-1.5t}cos(\omega t)

b)

m = 48lb

c)

b = 144.76lb

Explanation:

The general equation of a damping oscillate motion is given by:

u(t)=u_oe^{-\frac{b}{2m}t}cos(\omega t-\alpha)    (1)

uo: initial position

m: mass of the block

b: damping coefficient

w: angular frequency

α: initial phase

a. With the information given in the statement you replace the values of the parameters in (1). But first, you calculate the constant b by using the information about the viscous resistance force:

|F_{vis}|=bv\\\\b=\frac{|F_{vis}|}{v}\\\\|F_{vis}|=27lbs=27*32.17ft.lb/s^2=868.59ft.lb/s^2\\\\b=\frac{868.59}{6}lb/s=144.76lb/s

Then, you obtain by replacing in (1):

6in = 0.499 ft

u(t)=0.499ft.e^{-\frac{144.76lb/s}{2(48lb)}t}cos(\omega t)\\\\u(t)=0.499ft.e^{-1.5t}cos(\omega t)

b.

mass, m = 48lb

c.

b = 144.76 lb/s

8 0
3 years ago
Sue and Jenny kick a soccer ball at exactly the same time. Sue’s foot exerts a force of 52.6 N to the north. Jenny’s foot exerts
Karolina [17]
Using the addition of forces using right angled triangles. The resultant force sqaured. = 112.8 sqaured + 52.6 squared. So resultant force sqaured is 15490.6. So the resultant force is the sqaure root of this which is 124N to 3 significant figures
6 0
3 years ago
Read 2 more answers
A thin-walled cylindrical pressure vessel is subjected to an internal gauge pressure, p=75 psip=75 psi. It had a wall thickness
Mekhanik [1.2K]

To solve this problem we must apply the concept related to the longitudinal effort and the effort of the hoop. The effort of the hoop is given as

\sigma_h = \frac{Pd}{2t}

Here,

P = Pressure

d = Diameter

t = Thickness

At the same time the longitudinal stress is given as,

\sigma_l = \frac{Pd}{4t}

The letters have the same meaning as before.

Then he hoop stress would be,

\sigma_h = \frac{Pd}{2t}

\sigma_h = \frac{75 \times 8}{2\times 0.25}

\sigma_h = 1200psi

And the longitudinal stress would be

\sigma_l = \frac{Pd}{4t}

\sigma_l = \frac{75\times 8}{4\times 0.25}

\sigma_l = 600Psi

The Mohr's circle is attached in a image to find the maximum shear stress, which is given as

\tau_{max} = \frac{\sigma_h}{2}

\tau_{max} = \frac{1200}{2}

\tau_{max} = 600Psi

Therefore the maximum shear stress in the pressure vessel when it is subjected to this pressure is 600Psi

6 0
3 years ago
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